Power Inductor in Buck Converter Experiment

Objective

The goal of this experiment is to analyze the role of a power inductor in a buck converter and to measure its effect on voltage regulation and output current in a DC-DC conversion circuit.

Materials Needed

Theory

A buck converter is a type of DC-DC converter that steps down a higher input voltage to a lower output voltage. It achieves this by using a combination of a switching transistor, diode, capacitor, and a power inductor. The inductor plays a critical role in smoothing out the current and reducing voltage ripple.

In this experiment, we will focus on the inductor's effect on the overall performance of the buck converter by measuring output voltage, current ripple, and the efficiency of the converter with different inductance values.

Steps

  1. Set Up the Buck Converter Circuit

    Use a standard buck converter module or assemble your own on a breadboard. The circuit should include a switching transistor, diode, capacitor, and power inductor. Refer to the buck converter circuit schematic for guidance.

    Connect the input of the buck converter to the DC power supply (e.g., 12V) and set a suitable output voltage (e.g., 5V) using the feedback resistor network.

  2. Install the First Inductor

    Begin with a power inductor of a specific value (e.g., 10µH) in the buck converter circuit. Make sure the inductor is rated for the current the converter is expected to handle.

  3. Measure Output Voltage and Ripple

    Use the oscilloscope to measure the output voltage and observe the voltage ripple across the load. Record the results for analysis later.

    Also measure the input and output currents using a multimeter to observe how the inductor affects the current through the load.

  4. Replace with Other Inductor Values

    Switch out the 10µH inductor for inductors with higher values (e.g., 22µH, 47µH). For each inductor, repeat the voltage and current measurements and observe the changes in the ripple and stability of the output.

  5. Analyze Efficiency

    Calculate the efficiency of the buck converter using the following formula:

    Efficiency (%) = (Pout / Pin) × 100

    Where Pout is the power delivered to the load (Vout × Iout), and Pin is the power drawn from the input supply (Vin × Iin).

Example Calculation

Assume the following values for a 10µH inductor:

The input power is:

Pin = Vin × Iin = 12V × 0.5A = 6W

The output power is:

Pout = Vout × Iout = 5V × 1A = 5W

The efficiency is therefore:

Efficiency = (5W / 6W) × 100 = 83.33%

Repeat this calculation for each inductor value and observe how the inductance affects the converter's efficiency and ripple.

Conclusion

In this experiment, we demonstrated the critical role of the power inductor in a buck converter's performance. The inductance value affects the voltage ripple, current stability, and efficiency of the converter. Larger inductors generally result in lower current ripple and higher efficiency, but they also slow down the converter's transient response. The choice of the inductor value depends on the application requirements, such as desired ripple and efficiency trade-offs.